# Find the flux of the curl of field F through the shell S. F = (e^x)i + (e^y)j + 2xyk S is the...

## Question:

Find the flux of the curl of field F through the shell S.

{eq}F = (e^x)i + (e^y)j + 2xyk {/eq}

S is the portion of the paraboloid {eq}2 - x^2 - y^2= z {/eq} that lies above the xy-plane.

## Vector Calculus

This problem is an application of vector calculus. There are various methods to solve this problem but we will use a more efficient one that will make use of some theorems of vector calculus. Once we have a proper relationship established we will calculate the flux.

We know that with the help of Stoke's theorem we can directly calculate the flux through a shell across the defined surface as:

{eq}\displaystyle \oint F.dr {/eq}

We have,

{eq}\displaystyle F=(e^x)\hat i+(e^y)\hat j+2xy \hat k {/eq}

The surface over which we have to calculate the curl is:

{eq}\displaystyle S:2-x^2-y^2=z {/eq} and that lies above the {eq}xy {/eq}-plane.

We can parametrize the equation as:

{eq}\displaystyle x=\sqrt 2 \cos t {/eq}

{eq}\displaystyle y=\sqrt 2 \sin t {/eq}

{eq}\displaystyle z=0 {/eq}

Plugging in the values of F and dr, we get,

{eq}\displaystyle \text{Curl}=\oint F.dr=\int_{0}^{\frac{\pi}{2}}\left ( e^{\sqrt 2\cos t}, e^{\sqrt 2\sin t}, 8\sin t\cos t \right ).\left ( -\sqrt 2 \sin t, \sqrt2 \cos t, 0 \right )dt {/eq}

{eq}\displaystyle \oint F.dr=\int_{0}^{\frac{\pi}{2}}\left ( e^{\sqrt 2\cos t}, e^{\sqrt 2\sin t}, 8\sin t\cos t \right ).\left ( -\sqrt 2 \sin t, \sqrt2 \cos t, 0 \right )dt {/eq}

{eq}\displaystyle \oint F.dr=\int_{0}^{\frac{\pi}{2}}\left ( \left ( -\sqrt 2\sin t \right )e^{\sqrt 2\cos t}+\left ( \sqrt 2\cos t \right )e^{\sqrt 2\sin t} \right )dt {/eq}

On integration, we get,

{eq}\displaystyle \oint F.dr=\left [ e^{\sqrt 2\cos t}+e^{\sqrt 2\sin t} \right ]_{0}^{\frac{\pi}{2}} {/eq}

{eq}\displaystyle \boxed{\displaystyle \oint F.dr=\left [ e^{\sqrt 2}-e^{\sqrt 2} \right ]=0} {/eq}

So, the flux of the curl of field F through the shell 0. 